Question 649476: find k so that y=kx-4 and y=3x^2+x+2 intersect twice
Answer by htmentor(1343)   (Show Source):
You can put this solution on YOUR website! find k so that y=kx-4 and y=3x^2+x+2 intersect twice
Setting the y-values equal, we have:
kx - 4 = 3x^2 + x + 2
Collect terms and set=0:
3x^2 + (1-k)x + 6 = 0
We can solve for x using the quadratic formula:
x = (-(1-k) +- sqrt((1-k)^2 - 72))/6
For there to be two solutions, the discriminant must be >0:
(1-k)^2 - 72 > 0
(1-k)^2 > 72
1-k > +/- sqrt(72)
Solving the inequality gives the two conditions on k:
k < 1 - sqrt(72)
k > 1 + sqrt(72)
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